Rather than me setting exercises for others to do, I'd like people to give me some challenging questions It's a thread where people can share nice problems, but I'd like the benefit of seeing if I can get to the result myself too.
By challenging questions I mean ones which require insight and clever tricks, or ones with nice results, rather than something that requires a lot of knowledge and can be done by appyling heavy theorems.
Content-wise they can contain anything that doesn't assume too much university-level knowledge, although I would like to learn some things along the way so don't hold back.
I'm mainly looking at people like Jane, mathsy and Ricky, but anyone else is of course welcome
Very well, heres a simple question for you.
Find all integer solutionsto the congruence equation
The solution uses material you should be familiar with (because Ive seen you discuss it on the forum). Let me know if you need a hint.
It would be helpful if we could multiply by a number n such that 13n is 1 more than a mutliple of 29. The smallest such number is 9. Mutliplying by 9 gives:
Thank you Jane
Last edited by Daniel123 (2008-11-20 11:44:17)
Heres a VERY VERY VERY easy sum:
If Peter is 4 times the age of Adam and Tom is ½ more than the age of Peter and the total sum of their ages is 47 then how old will each boy be?
TELL ME IF IT MAKES SENSE LOL
Writing the information out algebraically:
P = 4A
T = 1.5P
From here it is straightforward to solve.
Well the first part comes out from the sum of a geometric series:
Induction proves the second pretty quickly:
Obviously the first part could also be done (and more quickly) by induction, but I feel induction gets rid of the 'why' behind it. I'm sure you're going to show me a nicer way to do the second part
Last edited by Daniel123 (2008-12-25 00:00:29)
Well, you can do both parts in a nice way, actually.
All we hv to do is to prove that 9k^2 - 1 is divisible by 8.
=(3k+1)(3k-1) this reduces down to your other question -- product of 2 consecutive even nos is div by 8.......